Question: How much heat, in joules and in calories, is required to heat a 28.4-g (1-oz) ice cube from −23.0 °C to −1.0 °C?

The heat required to raise the temperature of a substance is given by the formula Q = C × m ×∆ T,where “C” is the specific heat of the substance, “m” is the mass of the substance, and “∆T” is the change in the temperature of the substance.

We know from the given details that:

C = 2.093 J/g °C(Table 5.1)

m = 28.4 g

\(\Delta \)T = change in temperature =\({T_{final}} - {T_{initial}} = \) -1\(^\circ C\) – ( - 23\(^0C\) ) = 22\(^0C\)

By putting the values above in the equation Q = C × m × ∆ T, we get

Q = 2.093\( \times \)28.4\( \times \)22 = 130.77 J.

The heat required to raise the temperature of 28.4 g of ice cube = 130.77 J

Also,

1 calorie = 4.184 J.

The heat required to raise the temperature in calories =\(\frac{{130.77}}{{4.186}} = 31.24cal\)